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Isosceles triangle and point P on the circumcircle

Source: JBMO 2002, Problem 1

October 29, 2005
geometrycircumcircletrigonometrytrig identitiesLaw of Cosines

Problem Statement

The triangle ABCABC has CA=CBCA = CB. PP is a point on the circumcircle between AA and BB (and on the opposite side of the line ABAB to CC). DD is the foot of the perpendicular from CC to PBPB. Show that PA+PB=2ā‹…PDPA + PB = 2 \cdot PD.
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